NCERT Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise – Straight Lines covers advanced problems based on all the concepts taught in Chapter 9. This exercise includes questions on finding equations of lines, distance of a point from a line, angle between two lines, and conditions for parallelism and perpendicularity.
Prepared according to the latest CBSE Class 11 Maths syllabus 2025-26, this Miscellaneous Exercise helps students revise and strengthen their understanding of straight lines in coordinate geometry. The problems are designed to test higher-order thinking and application skills, making them important for board exams and competitive exams like JEE.
NCERT Solutions for Class 11 Maths – Chapter 9 Straight Lines (Miscellaneous Exercise) 2025-26
Q.
Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.
Q.
Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x –x1) + B (y – y1) = 0.
Q.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0.
Q.
Q.
Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
Q.
Q.
Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l (x + y) + p = 0 and l (x + y) – r = 0.
Q.
Find equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).
Q.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Q.
Q.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y +19 = 0. at right angle. Find the value of h.
Q.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Q.
By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.
Q.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
Q.
Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Q.
The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.
Q.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2.
Q.
In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.
Q.
Q.
Find the equation of the line through the intersection of lines 3x + 4y = 7 and x – y + 2 = 0 and whose slope is 5.
Q.
Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0
Q.
Find the equation of the line through the intersection of the lines 2x + 3y – 4 = 0 and x – 5y = 7 that has its x-intercept equal to – 4.
Q.
If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2 (c3 – c1) + m3(c1– c2) = 0.
Q.
Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.
Q.
Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
Q.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Q.
Q.
Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
Q.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Q.
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
Q.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Q.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Q.
Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Q.
Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
Q.
Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).
Q.
Q.
A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).
Q.
Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?
Q.
Q.
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:
1. Write the equations for the x-and y-axes.
2.
3. Passing through (0, 0) with slope m.
4.
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
7. Passing through the points (–1, 1) and (2,– 4).
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.
Q.
The vertices of Δ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
Q.
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Q.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
Q.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Q.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Q.
Q.
The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.
Q.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.
Q.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?
Q.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
NCERT Solutions for Class 11 Maths – Chapter 9 Straight Lines (Miscellaneous Exercise) 2025-26
The solutions are explained in a clear, step-by-step format to help students understand each concept and solve problems with confidence.
Q1. Find the values of k for which the line (k−3)x − (4−k²)y + k² − 7k + 6 = 0 is: (a) Parallel to the x-axis, (b) Parallel to the y-axis, (c) Passing through the origin.
Ans:
(a) Parallel to x-axis: Coefficient of x = 0 k − 3 = 0 → k = 3
(b) Parallel to y-axis: Coefficient of y = 0 4 − k² = 0 → k² = 4 → k = ±2
(c) Passing through origin: Substituting (0,0): k² − 7k + 6 = 0 → (k−1)(k−6) = 0 → k = 1 or k = 6
Q2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.
Ans: √3x + y + 2 = 0 → −√3x − y = 2
Dividing by √[(−√3)² + (−1)²] = √(3+1) = 2:
−(√3/2)x − (1/2)y = 1
Comparing with x cos θ + y sin θ = p:
cos θ = −√3/2, sin θ = −1/2
θ = 210° (or 7π/6), p = 1
Q3. Find the equations of the lines, which cut off intercepts on the axes whose sum and product are 1 and −6 respectively.
Ans: Let intercepts be a and b. a + b = 1 and ab = −6
So a and b are roots of: t² − t − 6 = 0 → (t−3)(t+2) = 0
→ t = 3 or t = −2
Case I: a = 3, b = −2 → x/3 + y/(−2) = 1 → 2x − 3y = 6
Case II: a = −2, b = 3 → x/(−2) + y/3 = 1 → −3x + 2y = 6 → 3x − 2y + 6 = 0
Q4. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?
Ans: Line: x/3 + y/4 = 1 → 4x + 3y − 12 = 0
Point on y-axis: (0, b)
Distance = |4(0) + 3b − 12| / √(16+9) = |3b−12|/5 = 4
|3b − 12| = 20
3b − 12 = 20 → b = 32/3 → (0, 32/3)
3b − 12 = −20 → b = −8/3 → (0, −8/3)
Q5. Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos φ, sin φ).
Ans: Equation of line through (cos θ, sin θ) and (cos φ, sin φ):
After simplification, the distance from origin:
d = |cos((θ−φ)/2)| = cos((θ−φ)/2)
(Since the perpendicular distance = 1 / |sec((θ−φ)/2)| = cos((θ−φ)/2))
Q6. Find the equation of the line parallel to y-axis and drawn through the point of intersection of x − 7y + 5 = 0 and 3x + y = 0.
Ans: Solving x − 7y + 5 = 0 and 3x + y = 0:
From 3x + y = 0 → y = −3x
x − 7(−3x) + 5 = 0 → x + 21x = −5 → 22x = −5 → x = −5/22
A line parallel to y-axis: x = constant
x = −5/22
Q7. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.
Ans: Line x/4 + y/6 = 1 meets y-axis at x = 0:
0 + y/6 = 1 → y = 6 → Point = (0, 6)
Slope of given line: 6x + 4y = 24 → slope = −6/4 = −3/2
Slope of perpendicular = 2/3
Equation: y − 6 = (2/3)(x − 0)
2x − 3y + 18 = 0
Q8. Find the area of the triangle formed by the lines y − x = 0, x + y = 0 and x − k = 0.
Ans: Lines: y = x, y = −x, x = k
Vertices:
- y = x and y = −x → (0, 0)
- y = x and x = k → (k, k)
- y = −x and x = k → (k, −k)
Area = (1/2)|base × height| = (1/2)(2k)(k) = k²
Q9. Find the value of p so that the three lines 3x + y − 2 = 0, px + 2y − 3 = 0 and 2x − y − 3 = 0 may meet at a point.
Ans: Solving 3x + y − 2 = 0 and 2x − y − 3 = 0:
Adding: 5x = 5 → x = 1, y = −1 → Point (1, −1)
Substituting in px + 2y − 3 = 0:
p(1) + 2(−1) − 3 = 0 → p − 5 = 0 → p = 5
Q10. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.
Ans: For concurrent lines, point of intersection of first two lies on third.
From y = m₁x + c₁ and y = m₂x + c₂:
x = (c₁ − c₂)/(m₂ − m₁), y = (m₂c₁ − m₁c₂)/(m₂ − m₁)
Substituting in y = m₃x + c₃ and simplifying:
m₂c₁ − m₁c₂ = m₃(c₁ − c₂) + c₃(m₂ − m₁)
Rearranging: m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0 (Proved)
Q11. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x − 2y = 3.
Ans: Slope of x − 2y = 3 is m₁ = 1/2
tan 45° = |m − m₁|/|1 + mm₁| = 1
Case I: (m − 1/2)/(1 + m/2) = 1 → m − 1/2 = 1 + m/2 → m/2 = 3/2 → m = 3
Line: y − 2 = 3(x − 3) → 3x − y = 7
Case II: (m − 1/2)/(1 + m/2) = −1 → m − 1/2 = −1 − m/2 → 3m/2 = −1/2 → m = −1/3
Line: y − 2 = −(1/3)(x − 3) → x + 3y = 9
Q12. Find the equation of the line passing through the point of intersection of 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 that has equal intercepts on the axes.
Ans: Solving 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0:
From 2x − 3y + 1 = 0 → x = (3y−1)/2
4(3y−1)/2 + 7y − 3 = 0 → 6y − 2 + 7y − 3 = 0 → 13y = 5 → y = 5/13
x = (15/13 − 1)/2 = (2/13)/2 = 1/13
Point of intersection = (1/13, 5/13)
Equal intercepts → equation: x + y = a
1/13 + 5/13 = a → a = 6/13
x + y = 6/13 → 13x + 13y = 6
Q13. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (m ± tan θ)/(1 ∓ m tan θ).
Ans: Line through origin: y = Mx
Angle between y = Mx and y = mx + c:
tan θ = |(M − m)/(1 + Mm)|
Taking both cases: M = (m + tan θ)/(1 − m tan θ) and M = (m − tan θ)/(1 + m tan θ)
Since M = y/x:
y/x = (m ± tan θ)/(1 ∓ m tan θ) (Proved)
Q14. In what ratio, the line joining (−1, 1) and (5, 7) is divided by the line x + y = 4?
Ans: Let line x + y = 4 divide the join of (−1,1) and (5,7) in ratio k:1.
Point of division = ((5k−1)/(k+1), (7k+1)/(k+1))
Substituting in x + y = 4:
(5k−1 + 7k+1)/(k+1) = 4
12k/(k+1) = 4 → 12k = 4k + 4 → 8k = 4 → k = 1/2
Ratio = 1:2
Q15. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x − y = 0.
Ans: Line through (1,2) parallel to 2x − y = 0: slope = 2
Parametric form through (1,2): x = 1 + t cos α, y = 2 + t sin α where tan α = 2
Intersection with 4x + 7y + 5 = 0:
4(1 + t·1/√5) + 7(2 + t·2/√5) + 5 = 0
4 + 4t/√5 + 14 + 14t/√5 + 5 = 0
23 + 18t/√5 = 0 → t = −23√5/18
Distance = 23√5/18
Q16. Find the direction in which a straight line must be drawn through the point (−1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Ans: Let the line through (−1, 2) make angle θ with x-axis.
Parametric: x = −1 + 3 cos θ, y = 2 + 3 sin θ (distance = 3)
Point lies on x + y = 4:
(−1 + 3 cos θ) + (2 + 3 sin θ) = 4
1 + 3(cos θ + sin θ) = 4
cos θ + sin θ = 1
√2 sin(θ + 45°) = 1 → sin(θ + 45°) = 1/√2
θ + 45° = 45° or 135° → θ = 0° or θ = 90°
The line must be drawn in the direction of 60° with the x-axis (i.e., parallel to y-axis, θ = 90°) or along x-axis (θ = 0°).
Q17. The hypotenuse of a right triangle is along the line 3x − 4y = 5. The right angle vertex is at (1, 1). Find the two legs of the triangle.
Ans: Foot of perpendicular from (1,1) on 3x − 4y = 5:
Foot = (1 + 3·(3−4−5)/25, 1 − 4·(3−4−5)/25) = (1 + 3·(−6)/25, 1 + 4·6/25)
= (1 − 18/25, 1 + 24/25) = (7/25, 49/25)
Leg 1: Line joining (1,1) and (7/25, 49/25) — this is perpendicular to hypotenuse.
Slope = (49/25 − 1)/(7/25 − 1) = (24/25)/(−18/25) = −4/3
Wait — slope of 3x−4y=5 is 3/4, so perpendicular slope = −4/3.
Line through (1,1) with slope −4/3: 4x + 3y = 7
The two legs are: 4x + 3y = 7 and the leg along the hypotenuse direction through (1,1): 3x − 4y + 1 = 0
Q18. Find the image of the point (3, 8) with respect to the line x + 3y = 7, assuming the line to be a plane mirror.
Ans: Let image of (3, 8) be (h, k).
Midpoint of (3,8) and (h,k) lies on x + 3y = 7:
(3+h)/2 + 3(8+k)/2 = 7 → 3+h + 24+3k = 14 → h + 3k = −13 ...(i)
Line joining (3,8) and (h,k) is perpendicular to x + 3y = 7 (slope = −1/3):
Slope of (h,k) to (3,8) = (k−8)/(h−3) = 3
k − 8 = 3h − 9 → 3h − k = 1 ...(ii)
Solving (i) and (ii): h + 3(3h−1) = −13 → 10h = −10 → h = −1, k = −4
Image = (−1, −4)
Q19. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
Ans: m₁ = 3 (slope of y = 3x+1), m₂ = 1/2 (slope of 2y = x+3)
tan α₁ = |(m−3)/(1+3m)|, tan α₂ = |(m−1/2)/(1+m/2)|
Setting tan α₁ = tan α₂:
|(m−3)/(1+3m)| = |(2m−1)/(2+m)|
Case I: (m−3)(2+m) = (2m−1)(1+3m)
2m + m² − 6 − 3m = 2m + 6m² − 1 − 3m
m² − m − 6 = 6m² − m − 1 → 5m² = −5 → No real solution
Case II: (m−3)(2+m) = −(2m−1)(1+3m)
m² − m − 6 = −(6m² − m − 1) → 7m² = 5 → m = ±√(5/7) = ±(1/7)√35
Q20. If the sum of the perpendicular distances of a variable point P(x, y) from the lines x + y − 5 = 0 and 3x − 2y + 7 = 0 is always 10, show that P must move on a line.
Ans: Distance from x + y − 5 = 0: |x + y − 5|/√2
Distance from 3x − 2y + 7 = 0: |3x − 2y + 7|/√13
Sum = 10 (constant)
|x + y − 5|/√2 + |3x − 2y + 7|/√13 = 10
For a specific sign combination (taking positive values):
(x + y − 5)/√2 + (3x − 2y + 7)/√13 = 10
This is a linear equation in x and y → P moves on a straight line. (Proved)
Q21. Find the equation of the line which is equidistant from parallel lines 9x + 6y − 7 = 0 and 3x + 2y + 6 = 0.
Ans: Rewriting: 9x + 6y − 7 = 0 and 9x + 6y + 18 = 0
A line equidistant from both parallel lines passes through the midpoint of perpendicular distance.
Distance between lines = |−7 − 18|/√(81+36) = 25/√117
The equidistant line: 9x + 6y + (−7+18)/2 = 0
9x + 6y + 11/2 = 0
18x + 12y + 11 = 0
Q22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Ans: Image of (1, 2) in x-axis = (1, −2)
Line joining (1, −2) and (5, 3):
Slope = (3−(−2))/(5−1) = 5/4
Equation: y + 2 = (5/4)(x − 1) → 4y + 8 = 5x − 5 → 5x − 4y = 13
At y = 0 (x-axis): 5x = 13 → x = 13/5
A = (13/5, 0)
Q23. Prove that the product of the lengths of the perpendiculars drawn from the points (√(a²−b²), 0) and (−√(a²−b²), 0) to the line x/a · cos θ + y/b · sin θ = 1 is b².
Ans: Line: bx cos θ + ay sin θ − ab = 0
Distance from (√(a²−b²), 0):
d₁ = |b·√(a²−b²)·cos θ − ab| / √(b²cos²θ + a²sin²θ)
Distance from (−√(a²−b²), 0):
d₂ = |−b·√(a²−b²)·cos θ − ab| / √(b²cos²θ + a²sin²θ)
d₁ × d₂ = |b²(a²−b²)cos²θ − a²b²| / (b²cos²θ + a²sin²θ)
= b²|a²cos²θ − b²cos²θ − a²| / (b²cos²θ + a²sin²θ)
= b²|−a²sin²θ − b²cos²θ| / (b²cos²θ + a²sin²θ)
= b²(a²sin²θ + b²cos²θ) / (b²cos²θ + a²sin²θ) = b² (Proved)
Q24. A person standing at the junction (crossing) of two straight paths represented by the equations 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 wants to reach the path whose equation is 6x − 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Ans: Intersection of 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0:
From 2x − 3y = −4 ...(i) and 3x + 4y = 5 ...(ii)
Multiply (i) by 4 and (ii) by 3: 8x − 12y = −16 and 9x + 12y = 15
Adding: 17x = −1 → x = −1/17, y = (−4 + 2/17)/−3 = 22/17
Point = (−1/17, 22/17)
Shortest path = perpendicular from this point to 6x − 7y + 8 = 0
Slope of 6x − 7y + 8 = 0 is 6/7 → perpendicular slope = −7/6
Equation: y − 22/17 = −7/6(x + 1/17)
Multiplying through: 119x + 102y = 125
FAQs – Class 11 Maths Chapter 9 Miscellaneous Exercise Straight Lines
Q1. What topics are covered in the Miscellaneous Exercise of Chapter 9? The Miscellaneous Exercise covers all topics of Chapter 9 including equations of lines in various forms, distance of a point from a line, angle between two lines, concurrent lines, image of a point in a line, and equidistant lines.
Q2. What is the formula for the distance of a point from a line? The distance of point (x₁, y₁) from line Ax + By + C = 0 is: d = |Ax₁ + By₁ + C| / √(A² + B²)
Q3. How do we find the angle between two lines? If slopes are m₁ and m₂, the angle θ between them is: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
Q4. What is the condition for three lines to be concurrent? Three lines are concurrent if the point of intersection of any two lines satisfies the equation of the third line.
Q5. Why is the Miscellaneous Exercise important for board exams? This exercise contains higher-order application problems that are frequently asked in CBSE board exams and JEE. It tests conceptual understanding and the ability to apply multiple formulas together.
Q6. How can students prepare effectively for this exercise? Students should thoroughly revise all forms of equations of a line, key formulas like distance formula, angle formula, and practice all 24 questions with step-by-step solutions for best results.